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Polynomials: Number of Roots

  1. Jun 26, 2010 #1

    jgens

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    I'm trying to prove that a polynomial function of degree [itex]n[/itex] has at most [itex]n[/itex] roots. I was thinking that I could accomplish this by induction on the degree of the polynomial but I wanted to make sure that this would work first. If someone could let me know if this approach will work, I would appreciate it. Thanks!
     
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  3. Jun 26, 2010 #2

    Office_Shredder

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    Why don't you give it a shot and see what happens. Induction is probably the most common way to prove this
     
  4. Jun 27, 2010 #3

    mathman

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    One approach is going backwards. Given one root for a polynomial of degree n, you can use synthetic division to get a polynomial of degree n-1 for the rest of the roots. Continue until the polynomial is down to 0=0. You can do this exactly n times.
     
  5. Jun 27, 2010 #4

    HallsofIvy

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    If [itex]x_1[/itex], [itex]x_2[/itex], ..., [itex]x_n[/itex] are the roots of a polynomial equation, then the polynomial is [itex]a(x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_n)[/itex] for some number a. Multiplied out, that has degree n.

    Notice that a polynomial of degree n has exactly n roots if we count "multiplicity" (if [itex]x_1= x_2[/itex] that still counts as "two roots") and if we count complex roots. Counting only distinct real roots, there still cannot be more than n roots.
     
  6. Jun 27, 2010 #5

    disregardthat

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    This is usually treated as a theorem of what OP's about to prove. I think it is at least most pedagogical to start out without the above, and rather let it be a corollary, thus keeping the 'natural order'.
     
  7. Jun 28, 2010 #6

    HallsofIvy

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    What you could do, then, is prove that a polynomial of degree n has at least one (complex) solution. Then show that dividing the polynomial by x-a where a is that root, gives a polynomial of degree n-1. Since a polynomial cannot have degree less than 0, there can be, at most, n roots.
     
  8. Jun 28, 2010 #7

    Hurkyl

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    An example of a ring where polynomials fail to have at most n roots is the ring of integers modulo 8. The polynomial f(x)=x2-1 has, in fact, 4 distinct roots: 1, 3, 5, and 7.

    Proving there are at most n roots over a field is not a trivial thing. It is not a hard, but it is subtle -- easy to overlook relevant points.
     
    Last edited: Jun 28, 2010
  9. Jun 28, 2010 #8

    disregardthat

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    Is not the fact that it is an integral domain sufficient?
     
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